幂级数求和

Let a power series $$S(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{4n+1},$$ then $1$ is the radius of convergence of $S$ .In fact $S(x)$ convergens for each $x\in[-1,1).$ My work is to find a closed form of this power series.$\\$

The following is my solution: $$S(x)=\sum_{n=1}^{\infty}\frac{(4n+1)x^{n}-4nx^{n}}{4n+1}=\sum_{n=1}^{\infty}x^{n}-4x\sum_{n=1}^{\infty}(\frac{x^{n}}{4n+1})^{'}.$$then we have $$S(x)+4x S^{'}(x)=\frac{x}{1-x}.(\text{a first-order linear differential equation })\Rightarrow $$ $$\mathbf{A.}\quad S(x)=\frac{1}{x^{\frac{1}{4}}}[C_{1}+\frac{\ln(1+x^{\frac{1}{4}})+2\arctan(x^{\frac{1}{4}})-\ln(1-x^{\frac{1}{4}})-4x^{\frac{1}{4}}}{4}](1>x>0);$$$$\mathbf{B.}\quad S(x)=-\frac{1}{8(-x)^{\frac{1}{4}}}\begin{Bmatrix} C_{2}+8(-x)^{\frac{1}{4}}\\+2\sqrt{2}\arctan[1-\sqrt{2}(-x)^{\frac{1}{4}}]\\-2\sqrt{2}\arctan[1+\sqrt{2}(-x)^{\frac{1}{4}}]\\+ \sqrt{2}\ln[1-\sqrt{2}(-x)^{\frac{1}{4}}+\sqrt{-x}]\\-\sqrt{2}\ln[1+\sqrt{2}(-x)^{\frac{1}{4}}+\sqrt{-x}] \end{Bmatrix}\quad(-1\leq x<0).$$ Since $S(0)=0$ and $S(x)$ is continuous at $x=0$ , I conclude that $C_{1}=C_{2}=0.\diamondsuit$